Hi,

I have the equation:

$\displaystyle \begin{cases}y^{\prime\prime} + 3y^{\prime} + 2y = sin~x \\

y(0) = 1 \\

y^{\prime}(0) = 1 \end{cases}$

The auxiliary equation $\displaystyle t^2 + 3t + 2 = 0$ has the solutions

$\displaystyle x_{1}=-1~~ x_{2}=-2$

So the general solution would look like this

$\displaystyle Ae^{-t}+Be^{-2t}$

Now, let $\displaystyle \begin{cases}y = A~Cos(t)+B~Sin(t) \\

y^{\prime} = -A~Sin(t)+B~Cos(t) \\

y^{\prime\prime} = -A~Cos(t)-B~Sin(t) \end{cases}$

gives $\displaystyle -A~Cos(t)-B~Sin(t) + 3\cdot(-A~Sin(t)+B~Cos(t)) + 2\cdot(A~Cos(t)+B~Sin(t))$

Can be simplified down to $\displaystyle 3A~Cos(t)-3B~Sin(t) + ASin(t) +B~Cos(t)$

This is where i am stuck, how do you combine this with the general solution?

//Thanks